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in Physics by (44.7k points)

100 g ice at 0°C placed in 100 g water at 100°C. The final temperature of the mixture will be......... (Latent heat of ice is 80 Cal/g, and specific heat of water is 1 Cal/g C°) 

(A) 10°C 

(B) 20°C 

(C) 30°C 

(D) 50°C

1 Answer

+1 vote
by (69.7k points)
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Best answer

The correct option is (A) 10°C.

Explanation:

Let the temperature of mixture = T 

∴ heat absorbed by ice = heat lost by water 

∴ heat required to melt ice + heat to reach T = heat lost by water. 

∴ mL + mC(T – O) = mC(100 – T) 

Given L = Latent heat of ice = 80 Cal/g C = 1 Cal/9°C 

∴ (100 × 80) + (100 × 1)T = 100 × 1(100 – T) 

∴ 8000 + 100T = 10000 – 100T 

∴ 200T = 2000 T = 10°C

by (10 points)
In such question ..didnt we take that full ice melts into water ..i mean why there is the need to use m×c×delta t for ice

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